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Q9  ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show
that \frac{AO}{BO} = \frac{CO}{DO}
 

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Draw a line EF passing through point O such that EO||CD\, \, and\, \, FO||CD

To prove : 

                \frac{AO}{BO} = \frac{CO}{DO}

In \triangle ADC, we have CD||EO

So, by using basic proportionality theorem, 

\frac{AE}{ED}=\frac{AO}{OC}........................................1

In \triangle ABD, we have AB||EO

So, by using basic proportionality theorem, 

\frac{DE}{EA}=\frac{OD}{BO}........................................2

Using equation 1 and 2, we get 

\frac{AO}{OC}=\frac{BO}{OD}

\Rightarrow \frac{AO}{BO} = \frac{CO}{DO}

Hence proved 

 

 

Posted by

seema garhwal

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